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Binary To Octal Conversion | C Program
Binary to Octal conversion in C
There are two approaches of converting an Binary Number into a Octal Number like
- Converting Binary to Decimal and then Decimal to Octal
- By grouping Binary Number into Groups of 3 digits/bits and then converting to Octal
We will cover both approaches in this post
Method 1:-
We will first convert the Binary number into Decimal and then convert the Decimal number into Octal
Make sure that you have gone through these approaches below –
Method 1 Code
Run
#include<stdio.h>
#include<math.h>
// function to convert binary to octal
void convert(long long num)
{
int octalDigit = 0, count = 1, i = 0, pos = 0;
int octalArray[32] = {0};
while(num != 0)
{
int digit = num % 10;
octalDigit += digit * pow(2, i);
i++;
num /= 10;
// placing current octalsum for 3 pair in array index position
octalArray[pos] = octalDigit;
// whenever we have read next 3 digits
// setting values to default
// increasing pos so next values can be placed at next array index
if(count % 3 == 0)
{
octalDigit = 0;
i = 0;
pos++;
}
count++;
}
// printing octal array in reverse order
for (int j = pos; j >= 0; j--)
printf("%d",octalArray[j]);
}
//main program
int main()
{
// long used rather than int to store large values
long long binary;
printf("Enter binary number: ");
scanf("%lld", &binary);
convert(binary);
return 0;
}
Output :
Enter binary number: 1010
Decimal : 10
Octal : 12
Method 2:-
Method 2 uses the concept of grouping 3 successive digits/bits of the binary number and calculating octal digits against each grouping
We use an additional array to store the octal digits at each index.
We will need to print the array in reverse to get actual octal equivalent.
Method 2 Code
Run
#include<stdio.h>
#include<math.h>
// function to convert binary to octal
void convert(long long num)
{
int octalDigit = 0, count = 1, i = 0, pos = 0;
int octalArray[32] = {0};
while(num != 0)
{
int digit = num % 10;
octalDigit += digit * pow(2, i);
i++;
num /= 10;
// placing current octalsum for 3 pair in array index position
octalArray[pos] = octalDigit;
// whenever we have read next 3 digits
// setting values to default
// increasing pos so next values can be placed at next array index
if(count % 3 == 0)
{
octalDigit = 0;
i = 0;
pos++;
}
count++;
}
// printing octal array in reverse order
for (int j = pos; j >= 0; j--)
printf("%d",octalArray[j]);
}
//main program
int main()
{
// long used rather than int to store large values
long long binary;
printf("Enter binary number: ");
scanf("%lld", &binary);
convert(binary);
return 0;
}
Output :
Enter binary number: 010101
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#include
#include
// function to convert binary to octal
void convert(long long num)
{
int octalDigit = 0, count = 1, i = 0, pos = 0;
int octalArray[32] = {0};
while(num != 0)
{
int digit = num % 10;
octalDigit += digit * pow(2, i);
i++;
num /= 10;
// placing current octalsum for 3 pair in array index position
octalArray[pos] = octalDigit;
// whenever we have read next 3 digits
// setting values to default
// increasing pos so next values can be placed at next array index
if(count % 3 == 0)
{
octalDigit = 0;
i = 0;
pos++;
}
count++;
}
// printing octal array in reverse order
for (int j = pos; j >= 0; j–)
printf(“%d”,octalArray[j]);
}
//main program
int main()
{
// long used rather than int to store large values
long long binary;
printf(“Enter binary number: “);
scanf(“%lld”, &binary);
convert(binary);
return 0;
}
#include
#include
long n;
int octalarray[32];
int binarytodecimal(int n)
{
int i=0,digit,decimal;
while(n!=0)
{
digit=n%10;
decimal+=digit*pow(2,i);
n/=10;
i++;
}
return decimal;
}
int decimaltooctal(int n)
{
int i=0,j;
int x= binarytodecimal(n);
while(x!=0)
{
octalarray[i]=x%8;
x/=8;
i++;
}
for(j=i-1;j>=0;j–)
{
printf(“%ld”,octalarray[j]);
}
}
int main()
{
printf(“enter the binary number”);
scanf(“%ld”,&n);
binarytodecimal(n);
decimaltooctal(n);
return 0;
}
#include
#include
int main()
{
int binary,rem,decimal=0,octal=0 ,weight=1,num;
printf(“enter the binary number :\n”);
scanf(“%d”,&binary);
num=binary;
while(num!= 0)
{
rem=num%10;
decimal=decimal+rem*weight;
weight=weight*2;
num=num/10;
}
weight=1;
rem =0;
while(decimal!= 0)
{
rem = decimal%8;
octal = octal+rem*weight;
weight = weight * 10;
decimal = decimal/8;
}
printf(“the binary number =%d and its octal number is=%d”,binary,octal);
getch();
return 0;
}
#include
int main()
{
int binary,decimal=0,octal=0,m=1,n=1;;
scanf(“%d”,&binary);
while(binary>0)
{
decimal=decimal+(binary%10)*m;
m=m*2;
binary=binary/10;
if(m==8)
{
octal=octal+n*decimal;n=n*10;m=1;decimal=0;
}
}
octal=octal+n*decimal;
printf(“%d”,octal);
return 0;
}
This code is same as binary to decimal
binary to octal is as follows:
// Online C compiler to run C program online
#include
int binary2octal(int n);
int binary2decimal(int n);
int main()
{
int number,octal;
printf(“enter binary number”);
scanf(“%d”,&number);
octal= binary2octal(number);
printf(“bin to octal of %d is %d”,number,octal);
}
int binary2octal(int number)
{
int octal=0,value,i=1,lastdigit;
while(number !=0)
{
lastdigit=number%1000;
value= binary2decimal(lastdigit);
octal=octal+value*i;
i=i*10;
number=number/1000;
} return octal;
}
int binary2decimal(int number)
{
int dn=0,base=1,rem;
while(number!=0)
{
rem=number%10;
dn=dn+rem*base;
base=base*2;
number=number/10;
}
return dn;
}
#include
#include
int convert(long long bin);
int main() {
long long bin;
printf(“Enter a binary number: “);
scanf(“%lld”, &bin);
printf(“%lld in binary = %d in octal”, bin, convert(bin));
return 0;
}
int convert(long long bin) {
int oct = 0, dec = 0, i = 0;
// converting binary to decimal
while (bin != 0) {
dec += (bin % 10) * pow(2, i);
++i;
bin /= 10;
}
i = 1;
// converting to decimal to octal
while (dec != 0) {
oct += (dec % 8) * i;
dec /= 8;
i *= 10;
}
return oct;
}
This code is wrong as it converts binary to decimal only
when input is 10101001 the above code prints 169 as its octal no.which is wrong.
definately this code is binary to decimal conversion
Sorry for the mistake guys, we’ll fix it up asap. Thank you
You have not fixed this code yet…!
#include
#include
int main()
{
int n,rem,d=0,b=0;
scanf(“%d”,&n);
while(n>0){
rem=n%10;
d=d+(pow(2,b)*rem);
b++;
n/=10;
}
int o=0,i=1;
while(d!=0){
o=o+((d%8)*i);
d=d/8;
i=i*10;
}
printf(“%d”,o);
return 0;
}